Speedboat Acceleration Problem: How Long to Reach the Buoy?
How long does it take the boat to reach the buoy?
Given data: A speedboat moving at 29.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.4 m/s2 by reducing the throttle.
Answer: t = 4.8 sec
Explanation:
From the question we are told that:
Velocity of speed boat V = 29.0 m/s
Distance to Marker d = 100 m
Acceleration a = -3.4 m/s2
Generally the Newton's 3rd motion equation is given as:
v^2 = u^2 + 2 * a * s
v^2 = 29^2 + 2 * (-3.4) * 100
v = √161
v = 12.68 m/s
Generally the Newton's first equation of motion is given as:
v = u + a * t
12.68 = 29 - 3.4 * t
12.68 - 29 = -3.4t
-16.32 = -3.4t
t = (-16.32) / (-3.4)
t = 4.8 sec
In this scenario, we are dealing with a speedboat that is decelerating as it approaches a buoy marker at a constant acceleration rate of -3.4 m/s2. The initial velocity of the boat is 29.0 m/s, and the distance to the marker is 100 m.
By using the equations of motion, we can calculate the time it takes for the boat to reach the buoy. First, we utilize the Newton's 3rd motion equation to find the final velocity of the boat, which turns out to be 12.68 m/s.
Next, applying the Newton's first equation of motion, we can determine the time taken for the boat to reach the buoy, which is 4.8 seconds. This means it will take the boat 4.8 seconds to reach the buoy from its initial velocity of 29.0 m/s with a constant deceleration rate of -3.4 m/s2.